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Use of japi.Util.immutableSeq(j.u.List) is inefficient because it iterates over the list to create a new Seq #1488

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pjfanning opened this issue Sep 20, 2024 · 0 comments
Open

Use of japi.Util.immutableSeq(j.u.List) is inefficient because it iterates over the list to create a new Seq #1488

pjfanning opened this issue Sep 20, 2024 · 0 comments

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@pjfanning
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pekko/actor/src/main/scala/org/apache/pekko/japi/JavaAPI.scala

Lines 269 to 281 in 22a7385

def immutableSeq[T](iterable: java.lang.Iterable[T]): immutable.Seq[T] =
iterable match {
case imm: immutable.Seq[_] => imm.asInstanceOf[immutable.Seq[T]]
case other =>
val i = other.iterator()
if (i.hasNext) {
val builder = new immutable.VectorBuilder[T]
while ({ builder += i.next(); i.hasNext }) ()
builder.result()
} else EmptyImmutableSeq
}

It's used in a lot of places internally but the most performance sensitive area is the pekko-stream code (eg Flow, Source, etc).
The obvious solution of using pekko.util.ccompat.JavaConverters to convert using .asScala.toSeq doesn't compile in Scala 2.12 because toSeq gives a scala.Seq instead of a scala.immutable.Seq.

We could use .asScala.toList but that is likely to be slower. We could also a new Util.immutableWrappingSeq function that has different implementations for Scala 2.12 (possibly delegate to existing immutableSeq) and Scala 2.13+ that uses .asScala.toSeq.
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@pjfanning